Golden ratio calculated from theory of modular units, or from Ramanujan’s class invariant?

Question

considering the following infinite product series:

$$\frac{1+\sqrt{5}}{2}= e^{\pi/6} \prod_{k=1}^\infty \frac{1+e^{-5(2k-1)\pi}}{1+e^{-(2k-1)\pi}}$$

My question: Would this formula fit within, or be obtainable by, the theory of modular units (by Kubert and Lang)? I am entirely unfamiliar with that level of math, but I was told that the theory produces algebraic values for special infinite series products at CM-points.

Of note: The infinite product series (which adjusts the logarithmic spiral at $30$ degrees) has as its repeating term: $(1 + e^{-k\pi})$, with $k$ representing the sequence of all odd integers excluding those which are divisible by five $(1,3,7,9,11,13,17,19,21,23,27,\ldots)$.

Further, the same repeating term with $k$ representing all odd integers adjusts the logarithmic spiral at $7.5$ degrees to the fourth root of $2$.

And further, the same repeating term with $k$ representing only the odd integers that are divisible by $5$, adjusts the logarithmic spiral at $37.5$ degrees to the value of the product of the golden ratio with the fourth root of $2$.

Thanks in advance for any comments that can explain the theory of modular units, or CM-points, in a simplified manner for me, or its relation to infinite product series such as this one.

Answer 1

It turns out that OP also posted this question on my blog page. And I provide the answer I posted there.

The formula can be expressed as $$\frac{1 + \sqrt{5}}{2} = e^{\pi/6}\prod_{k = 1}^{\infty}\frac{1 + e^{-5(2k - 1)\pi}}{1 + e^{-(2k - 1)\pi}} = \frac{2^{-1/4}e^{5\pi/24}}{2^{-1/4}e^{\pi/24}}\prod_{k = 1}^{\infty}\frac{1 + e^{-5(2k - 1)\pi}}{1 + e^{-(2k - 1)\pi}}$$ If we check the definition of Ramanujan's Class Invariant $$G_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}\prod_{k = 1}^{\infty}\left(1 + e^{-(2k - 1)\pi\sqrt{n}}\right)$$ then we find that the given formula says that $$\frac{1 + \sqrt{5}}{2} = \frac{G_{25}}{G_{1}}$$ I have already proved in my post that $G_{1} = 1, G_{25} = (1 + \sqrt{5})/2$. This establishes your formula.

I have given extensive details of the theory of Ramanujan's theta functions which is needed to understand the proofs in my blog posts. All of it is based on real-analysis / calculus and does not require the deep and complicated theory of Modular forms and Complex multiplication.

Update: Based on your comments below I think the formula $$e^{\pi/6} = 2^{3/8}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi})$$ is wrong. Let us put $q = e^{-2\pi}$ then we have $$\begin{aligned}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi}) &= \prod_{k = 1}^{\infty}(1 + q^{k})\\ &= \prod_{k = 1}^{\infty}\frac{1 - q^{2k}}{1 - q^{k}}\\ &= \prod_{k = 1}^{\infty}\frac{1}{1 - q^{2k - 1}}\\ &= \prod_{k = 1}^{\infty}\frac{1}{1 - e^{-2(2k - 1)\pi}}\\ &= \dfrac{1}{2^{1/4}e^{-\pi/12}g_{4}}\end{aligned}$$ We have the definition $$g_{n}= 2^{-1/4}e^{\pi\sqrt{n}/24}\prod_{k = 1}^{\infty}\left(1 - e^{-(2k - 1)\pi\sqrt{n}}\right)$$ so that $g_{4} = 1/g_{1} = (2\sqrt{2})^{1/12}$ which gives the product as $2^{-3/8}e^{\pi/12}$. Your formula should consist of $e^{\pi/12}$ and not $e^{\pi/6}$. Your second product $$e^{\pi/4} = \sqrt{2(1 + \sqrt{2})}\prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}$$ (which is correct) can also be calculated in similar fashion and it is linked with Ramanujan class invariants discussed in my blog post. The idea is to put $q = e^{-\pi\sqrt{n}}$ for some suitable value of $n$ and link the products with the invariants $G_{n}$ or $g_{n}$.